MCQ
The non-zero vectors are $\vec a , \vec b$ and $\vec c$ are related by $\vec a = 8\vec b$ and $\vec c = -7\vec b$. Then the angle between $\vec a$ and $\vec c$ is ............... $^\circ $
- A$0$
- B$45$
- C$90$
- ✓$180$
Given that, $\vec{a}=8 \vec{b}$ and $\vec{c}=-7 \vec{b}$
Let
$b=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$
Then,
$\vec{a}=8\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)$
And
$\vec{c}=-7\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)$
Consider that angle between $\vec{a}$ and $\vec{c}$ is $x$.
Thus, $\cos x=\frac{a \cdot c}{|a||c|}$
$\cos x=\frac{-56\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)}{7 \times 8 \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}} \cdot \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}$
$\cos x=-1$
Therefore,
$x=\pi$
hence angle between $\vec{a}$ and $\vec{c}$ is $\pi$
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